- Maximize $z(x,y)=x^2y^{\frac{1}{2}}$
- Subject to $7x+3y=30$
- Where $x,y\geq 0$
Antwoord 1 correct
Correct
Antwoord 2 optie
$z(2\frac{1}{7},5)=4\frac{29}{49}\sqrt{5}$
Antwoord 2 correct
Fout
Antwoord 3 optie
$z(0,0)=0$
Antwoord 3 correct
Fout
Antwoord 4 optie
None of the other answers is correct.
Antwoord 4 correct
Fout
Antwoord 1 optie
$z(3\frac{3}{7},2)=11\frac{37}{49}\sqrt{2}$
Antwoord 1 feedback
Correct: $\dfrac{z'_x(x,y)}{z'_y(x,y)}=\dfrac{2xy^{\frac{1}{2}}}{\frac{1}{2}x^2y^{-\frac{1}{2}}}=\dfrac{4y}{x}=\dfrac{7}{3}$ gives $x=\frac{12}{7}y$. Plugging this into the restriction gives $7\frac{12}{7}y+3y=30$, which results in $y=2$ with $x=\frac{12}{7}\cdot 2=3\frac{3}{7}$. $z(3\frac{3}{7},2)=11\frac{37}{49}\sqrt{2}$. We check the boundaries: $z(0,10)=0$ and $z(4\frac{2}{7},0)=0$. Hence, $z(3\frac{3}{7},2)=11\frac{37}{49}\sqrt{2}$ is the maximum.
Go on.
Go on.
Antwoord 2 feedback
Wrong: $\dfrac{2}{\frac{1}{2}}\neq 1$.
Try again.
Try again.
Antwoord 3 feedback
Wrong: $(x,y)=(0,0)$ does not satisfy the restriction.
See Constrained optimization functions of two variables.
See Constrained optimization functions of two variables.
Antwoord 4 feedback
Wrong: The correct answer is among them.
Try again.
Try again.